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Let M=[sin4θ1sin2θ1+cos2θcos4θ]=αI+βM1 where α=α(θ) and β=β(θ) are real numbers and I is the 2×2 identity matrix. If α= is the minimum of the set {α(θ):θ[0,2π]} and β= is the minimum of the set {β(θ):θ[0,2π]}
Then the value of α+β is:
(a) 3716
(b) 2916
(c) 3116
(d) 1716

Answer
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Hint: We have given M in such a way that M=[sin4θ1sin2θ1+cos2θcos4θ]=αI+βM1 and we have to find the value of α&β. For that, we are going to find the inverse of M. We know that inverse of a matrix M is equal to adj(M)|M|. In this formula, adj (M) is the adjoint of M which is calculated by taking the cofactor of matrix M then take the transpose of the matrix and |M| is the determinant of M. After calculating the inverse of M, substitute in the matrix given and also substitute the identity matrix and then play with the equation to see, what values of α&β will satisfy the equation. After that take the minimum value of α&β individually and then add them to get the required answer.

Complete step by step answer:
We have given the matrix M as follows:
M=[sin4θ1sin2θ1+cos2θcos4θ]=αI+βM1
To solve the above equation, we need the value of inverse of matrix M i.e. M1 so we are going to find the inverse of the matrix M.
We know the inverse of a matrix M is equal to:
adj(M)|M|
Now, we are going to find the adj (M) or adjoint of the matrix M by first of all taking the co – factor of M and then take the transpose of that matrix.
Finding the co – factor of matrix M as follows:
The cofactor of matrix M contains cofactor of first row and first column, first row and second column likewise you can write all the position of the elements. In the below, we are showing the cofactor at the position of first row and first column.
The cofactor of the element lies at ith row and jth column is given below:
Cij=(1)i+j(Mij)
In the above equation, Mij represents the minor corresponding to the position of i and j of the matrix.
The matrix M given as:
[sin4θ1sin2θ1+cos2θcos4θ]
Finding the cofactor of the position where sin4θ is placed so value of i and j are both equal to 1 we get,
C11=(1)1+1(M11)C11=(1)2(M11)C11=1(M11)
Now, we have to find the minor of the position (i, j) is equal to 1 which we have shown below:
Hide the elements of the first row and first column which we have shown below by red color and then the element remaining is the minor corresponding to the first row and first column.
[\colorredsin4\colorredθ\colorred\colorred1\colorred\colorredsin2\colorredθ\colorred1\colorred+\colorredcos2\colorredθcos4θ]
Hence, the value of M11=cos4θ
Similarly, we can find the other minors also so all the minors are:
M12=1+cos2θM21=1sin2θM22=sin4θ
Now, using these minors, we can find the co factors as follows:
C11=M11C11=cos4θ
C12=(1)1+2(M12)C12=1(1+cos2θ)
C21=(1)2+1(M21)C21=1(1sin2θ)
C22=(1)2+2(M22)C22=1(sin4θ)
Now, writing these cofactors in the matrix we get,
[C11C12C21C22]=[cos4θ1cos2θ1+sin2θsin4θ]
Taking the transpose of the above matrix by interchanging rows with columns we get,
[cos4θ1+sin2θ1cos2θsin4θ]
Hence, we got the adjoint of the matrix M as:
adj(M)=[cos4θ1+sin2θ1cos2θsin4θ]
Determinant of the matrix M is calculated as follows:
Matrix M is shown below;
[sin4θ1sin2θ1+cos2θcos4θ]
Taking determinant of the matrix we get,
sin4θcos4θ(1+cos2θ)(1)(1+sin2θ)=sin4θcos4θ+(1+cos2θ)(1+sin2θ)=sin4θcos4θ+1+cos2θ+sin2θ+sin2θcos2θ
We know that there is a trigonometric identity that cos2θ+sin2θ=1 so using this relation in the above we get,
1+1+sin4θcos4θ+sin2θcos2θ=2+sin4θcos4θ+sin2θcos2θ
Hence, we got the determinant value of matrix M as:
|M|=2+sin4θcos4θ+sin2θcos2θ
Now, the inverse of matrix M is equal to:
[cos4θ1+sin2θ1cos2θsin4θ]|M|
Multiplying |M| inside the matrix we get,
[cos4θ|M|1+sin2θ|M|1cos2θ|M|sin4θ|M|]
Substituting this inverse of M in the given matrix equation we get,
[sin4θ1sin2θ1+cos2θcos4θ]=α[1001]+β[cos4θ|M|1+sin2θ|M|1cos2θ|M|sin4θ|M|]
Solving the right hand side of the above equation we get,
[α00α]+[βcos4θ|M|β(1+sin2θ)|M|β(1cos2θ)|M|βsin4θ|M|]
Adding the above two matrices by adding the elements of first matrix with the corresponding row and column of the second matrix we get,
[α+βcos4θ|M|β(1+sin2θ)|M|β(1cos2θ)|M|α+βsin4θ|M|]
Now, equating the above matrix to the left hand side of the above equation we get,
[sin4θ1sin2θ1+cos2θcos4θ]=[α+βcos4θ|M|β(1+sin2θ)|M|β(1cos2θ)|M|α+βsin4θ|M|]
Above two matrices are equal when each of the elements of matrix on the left hand side is equal to the corresponding elements of the matrix on the right hand side of the above equation we get,
(1+sin2θ)=β(1+sin2θ)|M|1=β|M|β=1|M|
Substituting the value of |M| in the above equation we get,
β=(2+sin2θcos2θ+sin4θcos4θ)
Hence, we got the value of β as (2+sin2θcos2θ+sin4θcos4θ)
sin4θ=α+βcos4θ|M|
Substituting the value of β that we have solved above in the above equation we get,
sin4θ=α+|M|cos4θ|M|sin4θ=αcos4θsin4θ+cos4θ=α
Making the L.H.S of the above equation as the perfect square we get,
(sin2θ+cos2θ)22sin2θcos2θ=α12sin2θcos2θ=α
Hence, we got the value of α&β as follows:
α=12sin2θcos2θβ=(2+sin2θcos2θ+sin4θcos4θ)
Now, we have to find the minimum value of α&β which is denoted as α&β as follows:
α= Minimum of α
β= Minimum of β
Finding the minimum value of α we get,
α=12sin22θ4α=1sin22θ2
Now, the minimum value of the above expression occurs when sin22θ is maximum and we know that maximum value of sin2θ is 1 so the value of its square is also 1 substituting this maximum value of sin22θ in the above equation we get,
α=112α=12
Hence, we got the value of α=12
Finding the minimum value of β we get,
β=(2+sin2θcos2θ+sin4θcos4θ)
Substituting sin2θcos2θ=t in the above equation we get,
β=(2+t+t2)
The above equation will give minimum value when 2+t+t2 is having maximum value.
Now, we are going to find the maximum value of 2+t+t2 as follows:
y=2+t+t2
As we have taken sin2θcos2θ=t so rearranging the L.H.S of this equation we get,
sin22θ4
Now, the maximum value of this expression is 14 which is found when sin2θ=1 and so with the similar reason the minimum value of the above expression is 0.
And we need the maximum value of 2+t+t2 which is possible when t=14. Substituting this value of t in β=(2+t+t2) then we get,
β=(2+14+(14)2)β=(2+14+116)
β=(32+4+116)
β=(3716)
Hence, we got the minimum value of β=3716 or β=(3716).
Now, adding α&β we get,
α+β=123716α+β=83716=2916

So, the correct answer is “Option B”.

Note: The most possible mistake that could happen in this problem is the calculation mistake. While finding the cofactor of the matrix, there is a high chance of making calculation mistakes so be careful in your calculations.
While finding the minimum value of β which we have shown above as:
β=(2+sin2θcos2θ+sin4θcos4θ)
Do not jump to put the value of θ=π2 directly into the above equation. This is also a very common mistake that students make. In such type of problems, try to reduce this expression as much as you can like you can write sinθcosθ=sin2θ2 and then put the values of θ which can give the minimum value.

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